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Given fa (x)=1/|x-a|+3, calculate limit `int fa (x)` integral from 0 to 3 a --->`oo`

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greenbel | (Level 2) Honors

Posted June 25, 2013 at 1:25 PM via web

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Given fa (x)=1/|x-a|+3, calculate limit `int fa (x)`

integral from 0 to 3

a --->`oo`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 25, 2013 at 6:13 PM (Answer #1)

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You need to evaluate the limit of the given definite integral, hence, you need to evaluate the definite integral, such that:

`int_0^3 1/(|x - a| + 3) dx`

You need to use the absolute value definition, such that:

`|x - a| = {(x - a, x - a >= 0),(a - x, x - a < 0):}`

`|x - a| = {(x - a, x>=a),(a - x, x<a):}`

Since `x in [0,3]` and `a->oo` , then `|x - a| = a - x` , such that:

`int_0^3 1/(a - x + 3) dx = int_0^3 1/(a + 3 - x) dx`

You should come up with the substitution, such that:

`a + 3 - x = t => -dx = dt`

Changing the limits of integration yields:

`x = 0 => t = a + 3`

`x = 3 => t = a`

`int_(a+3)^a (-dt)/t = int_a^(a+3) (dt)/t = ln t|_a^(a+3)`

Using the fundamental theorem of calculus yields:

`int_(a+3)^a (-dt)/t = ln(a+3) - ln a`

Converting the difference of logarithms into a logarithm of quotient yields:

`int_(a+3)^a (-dt)/t = ln ((a+3)/a)`

You need to evaluate the limit such that:

`lim_(a->oo) int_0^3 1/(|x - a| + 3) dx = lim_(a->oo) ln ((a+3)/a)`

`lim_(a->oo) ln ((a+3)/a) = ln lim_(a->oo)((a+3)/a) = 1`

Hence, evaluating the given limit, under the given conditions, yields `lim_(a->oo) int_0^3 1/(|x - a| + 3) dx = 1.`

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